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Stels [109]
3 years ago
11

A racecar on a straight track, starting from rest*, steps on the

Physics
1 answer:
kirill115 [55]3 years ago
6 0
I think the answer is C
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You drop a 0.375 kg ball from a height of 1.37 m. It hits the ground and bounces up again to a height of 0.67 m. How much energy
Radda [10]

2.57 joule energy lose in the bounce .

<u>Explanation</u>:

when ball is the height of 1.37 m from the ground  it has some gravitational potential energy with respect to hits the ground  

Formula for gravitational potential energy given by  

Potential Energy = mgh

Where ,

m = mass  

g = acceleration due to gravity  

h = height

Potential energy when ball hits the ground

m= 0.375 kg

h = 1.37 m

g = 9.8 m/s²

Potential Energy = 0.375\times9.8\times1.37

Potential Energy = 5.03 joule

Potential energy when ball bounces up again

h= 0.67 m

Potential Energy = 0.375\times0.67\times9.8

Potential Energy = 2.46 joule

Energy loss = 5.03 - 2.46 = 2.57 joule

2.57 joule energy lose in the bounce

6 0
3 years ago
What are beats? A. periodic fluctuations in the velocity of sound waves B. periodic fluctuations in the wavelength of sound wave
777dan777 [17]

The answer is

C. periodic fluctuations in the intensity if sound waves.

3 0
3 years ago
Read 2 more answers
What unit is used to count atoms and molecules?
murzikaleks [220]

the answer in my opinion would be A

7 0
3 years ago
Determine the kinetic energy of a 1000 kg roller coaster car that is moving with speed of 40.0 m/s​
ololo11 [35]

Answer:

KE=800,000

Explanation:

The formula for kinetic energy is KE=1/2mv^2 or Kinetic Energy= 0.5*mass*velocity^2

so 1000 is the mass and 40 is the velocity

KE=0.5*1000*40^2

KE=0.5*1,000*1,600

KE=800,000 Joules

8 0
3 years ago
A cylinder of diameter 100 mm rolls from restdown a 5 m long ramp and its center of mass is moving with velocity 2 m/s at the bo
RoseWind [281]

Answer:

(a): a = 0.4m/s²

(b): α = 8 radians/s²

Explanation:

First we propose an equation to determine the linear acceleration and an equation to determine the space traveled in the ramp (5m):

a= (Vf-Vi)/t = (2m/s)/t

a: linear acceleration.

Vf: speed at the end of the ramp.

Vi: speed at the beginning of the ramp (zero).

d= (1/2)×a×t² = 5m

d: distance of the ramp (5m).

We replace the first equation in the second to determine the travel time on the ramp:

d = 5m = (1/2)×( (2m/s)/t)×t² = (1m/s)×t ⇒ t = 5s

And the linear acceleration will be:

a = (2m/s)/5s = 0.4m/s²

Now we determine the perimeter of the cylinder to know the linear distance traveled on the ramp in a revolution:

perimeter = π×diameter = π×0.1m = 0.3142m

To determine the angular acceleration we divide the linear acceleration by the radius of the cylinder:

α = (0.4m/s²)/(0.05m) = 8 radians/s²

α: angular aceleration.

3 0
3 years ago
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